7 code shift I calculate is the complement 0000 0111, sign bit inversion is 1000 0111? But the book is 1000 0110?

7 code shift I calculate is the complement 0000 0111, sign bit inversion is 1000 0111? But the book is 1000 0110?

The original code of 7 is 00000 111, the reverse code is 01111000, the complement code is 01111001, and the shift code is 11111001

The absolute value of the difference between 2006 and 2005x is equal to 1

2006-2005 x = 1 or - 1
X = 1 or 2007 / 2005

If f (1 / x) = x ^ 2-5x, then f (x) =?

Let t = 1 / X. then x = 1 / T, = = = > F (T) = (1 / T) & sup2; - (5 / T). That is, f (x) = (1-5x) / X & sup2;

Simplify (Y-X) (z-x) / (x-2y + Z) (x + y-2z) + (Z-Y) (X-Y) / (x-2z + y) (y + z-2x) + (x-z) (Y-Z) / (y + z-2x) (x-2y + Z)

∵x-2y+z=（x-y）-（y-z）,x+y-2z=（y-z）-（z-x）,y+z-2x=（z-x）-（x-y）．
Let X-Y = a, Y-Z = B, z-x = C, then
The original formula = - AC / (a-b) (B-C) + - BA / (B-C) (C-A) + - CB / (C-A) (a-b)
=-ac(c-a)+ba(a-b)+bc(b-c)/(a-b)(b-c)(c-a)
=-ac²-a²c+ba²-b²a+b²c-bc² /(a-b)(b-c)(c-a)
=-c²(a-b)-c(a²-b²)+ab(a-b) /(a-b)(b-c)(c-a)
=-(a-b)(c²-ca-cd+ab)/(a-b)(b-c)(c-a)
=-(a-b)(c-a)(c-b)/(a-b)(b-c)(c-a)
=1.

In Cartesian coordinate system, given points a (4,0), B (0,3), if a right triangle is congruent with RT △ ABO, and they have a common edge, please write out the unknown vertex coordinates of the triangle (do not need to write the calculation process). (hint: consider the three cases of Ao, Bo, AB being the common edge respectively)

As shown in the figure, if AB is the common edge, there are three answers (72259625), (4,3), (2825, - 2125); if Bo is the common edge, there are two answers (- 4,3) and (- 4,0); if Ao is the common edge, there are two answers (0, - 3) and (4, - 3)

Solving equation (9 / 10) ^ x = 1 / 3 to find x
1. Such as the title
It's stuck
One more question,
2. A, B, C are positive real numbers, 3 ^ a = 4 ^ B = 6 ^ C to find the relationship between a, B, C
A、1/c=1/a+1/b
B、2/c=2/a+1/b
C、1/c=2/a+2/b
D、2/c=1/a+2/b
This The first question is to count, the answer is 11
Known condition: Lg3 = 0.4771

Take the logarithm xlg (9 / 10) = LG (1 / 3) x (2lg3-1) = - lg3x = Lg3 / (1-2lg3) 2, a, B, C are positive real numbers, 3 ^ a = 4 ^ B = 6 ^ C to find the relationship a, B, C, 1 / C = 1 / A + 1 / BB, 2 / C = 2 / A + 1 / BC, 1 / C = 2 / A + 2 / BD, 2 / C = 1 / A + 2 / B3 ^ a = 4 ^ B = 6 ^ C so take the logarithm aln3 = bln4 = cln6so aln3 = 2

Let a1a2a3 be a maximal independent group of a vector group, and B1 = a1 + A2 + a3, B2 = a1 + A2 + 2A3, B3 = a1 + 2A2 + 3a3
B1b2b3 is also the maximal independent group of the vector group

It is known that a 1, a 2, a 3 are the base vectors of vector group, and the rank of the vector group is 3
Because A1 = B1 + b2-b3, A2 = B1 + b3-2b2, A3 = b2-b1,
Therefore, A1, A2 and A3 can be expressed linearly by vectors B1, B2 and B3,
So B 1, B 2, B 3 are also the base vectors of the vector group, which are maximally linearly independent

If the cubic power of the square of (M + 2) x and the N-1 power of Y are the sixth power monomials of X and y, what conditions should m and N satisfy

If the N-1 power of (M + 2) &# 178; X & # 179; y is a binomial with respect to X and y, then the sum of the exponents of X and Y is equal to 6, and the coefficient (M + 2) &# 178; cannot be equal to 0, the exponent of x is 3, and the exponent of Y is (n-1)
3 + n-1 = 6 and M + 2 ≠ 0
m≠-2,n=4

Let the sequence {an} be an arithmetic sequence, the sum of the first n terms be Sn, A2 = 2, S5 = 15 are known, and the general term formula of {an} is obtained
(2) If BN = the nth power of an / 2, find the first n terms of BN and t

The tolerance of {an} {an} {{an}} tolerance is d1.s5 = a1.s5 = a1 + A2 + A2 + a3 + a3 + a3 + a3 + a3 + a3 + A5 = 5A3 = 15a3 = 15a3 = 3, we also know that A2 = 2D = a3-a2 = a3-a2-a2 = 3-2 = 3-2 = 1An = a1 + (n-1) d = A2 + (n-1) d = A2 + (n-1) d = A2 + (n-1) d = A2 + (n-1) d = A2 + (n-1 (n-1) d = A2 + (n = n = n = n2.bn = an / 2 / 2 \; = n / 2 8319; 8319; 8319; 8319; 8319; 8319; 8319; TN = 8319; 8319; TN = 8319; TN = 8319; TN = 8319; TN = 8319; TN = B1 and

If the function f (x) = loga (the square of X - ax + 3) (a > 0 and not equal to 1) satisfies the following conditions for any x1, x2
When x1

Loga (x ^ 2-ax + 3) = ln (x ^ 2-ax + 3) / (LN (a)) replace lower logarithm
f(x1)-f(x2）=[ln(x1^2-a*x1+3)-ln(x2^2-a*x2+3)]/ln(a)
Analysis of the situation
When 0